How much energy is required to raise the temperature of a 3 kg of iron from 20 degrees celsius to 25 degrees celsius?

Answer 1

6750 J

Answer 2

This is a specific heat problem. The specific heat of iron is 0.444 J/goC. The following are the formula and the variables needed to answer this question.

q = m x C x ΔT

q = amount of heat energy gained or lost by substance = ?

m = mass of sample in grams = 5.58kg = 5580g

C = heat capacity (J/goC) = 0.444 J/goC

Tf = final temperature = 1000.0oC

Ti = initial temperature = 20.0oC

ΔT = (Tf - Ti) = 980.0oC

Solution:

q = 5580g x 0.444 J/goC x 980.0oC = 2.43 x 106 Joules (2427969.6 rounded to three significant figures) You could also indicate it as 2430000 Joules.

Answer 3

You want to use the Equation Q=MS0T. The M stands for the mass, which is 28.2g. The S stands for the specific heat. Which is .449/gC for iron. You can find specific heats on a specific heat table. The 0 (which isn't an 0 but a triangle in the equation) stands for the change in temperature. This is what you are trying to find. So use (x-20C). It is X-20C because you are adding heat, so temperature increases as a result. Set this equal to Q, which stands for heat energy, 3.5KJ. You will want to convert this to Joules first though. So, 3.5*1000= 3500 joules.

Now, you have the equation 3500 joules=28.2g(.449/gC)(X-20C). The C cancels the C, and the g cancels the g, and you are left with joules as your only unit. Now multiply 28.2 by .449 to get 12.6618. Divide that by it self and you are left with X-20. The X is almost isolated. Since you divided one side by 12.6618, do it to the other side. 3500/12.6618=276.4219937. The resulting equation is 276.4219937=X-20. Add 20 to both sides to get 296.4219937. Round off significant figures and you get 296C. The last step is to convert this to K. So, add 273 to 296 to get 569K.

Answer 4

The specific heat capacity (C) of iron is 449J/kgoC. The equation needed to solve this problem is Q = m•C•ΔT, Q is the quantity of heat gained or lost, m is the mass of the object, C is the specific heat capacity of the material is composed of, and ΔT is the resulting temperature change of the object. Tfinal - Tinitial = ΔT.

Q = 5.58kg x 449J/kgoC x (1000oC - 20oC) = 2.455 x 106 Joules

Answer 5

q(joules) = mass(in grams)(specific heat)(change in temperature)

( 3 kg = 3000 grams )

q = (3000 grams)(0.44 J/gC)(25o C - 20o C)

= 7 X 103 Joules
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Answer 6

The specific heat capacity of iron is 0.45 J/g/deg. Therefore...

q = (m)(C)(∆T) where m=mass; C=specific heat capacity and ∆T=change in temp.

q = (25.0g)(0.45 J/g/deg)(10 deg)

q = 113 J = 0.113 kJ

Answer 7

The temperature at which the vapor pressure of a liquid had external pressure of 1 ATM it the

Answer 8

This heat is 112,4 joules.